Answer:
a) v = 5.88 m/s
b) v = 11.76 m/s
c) s = 1.764 m
d) s = 7.056 m
Explanation:
Given,
The initial velocity of the steel ball, u = o
The acceleration due to gravity, g = 9.8 m/s²
The equations of motion to find the final velocity for a body under free fall with an initial velocity, u = 0 is
v = u + at m/s
v = at m/s
a) At time t = 0.6 s
v = 9.8 x 0.6
= 5.88 m/s
The velocity of the ball 0.6 seconds after its release is, v = 5.88 m/s
b) At t = 1.2 s
v = 9.8 x 1.2
= 11.76 m/s
The velocity of the ball 1.2 seconds after its release is, v = 11.76 m/s
The distance traveled by the free falling body is given by the formula
s = ut + 1/2 at² m
s = 1/2 at² ∵ u = 0
c) At, t = 0.6 s
s = 1/2 x 9.8 x 0.6²
= 1.764 m
The distance ball fall in the first 0.6 seconds of its flight is, s = 1.764 m
d) At, t = 1.2 s
s = 1/2 x 9.8 x 1.2²
= 7.056 m
The distance ball fall in the first 1.2 seconds of its flight is, s = 7.056 m
